Question

An incompressible fluid flows steadily through a pipe that has a change in diameter. The fluid speed at a location where the pipe diameter is 8.8 cm is 2.4 m/s. Calculate the flow speed at a location where the diameter has narrowed to 5.80 cm

Answer 1

Answer:

The value is $v_2 = 5.53 \ m /s$

Explanation:

From the question we are told

  The pipe diameter at location 1 is  $d = 8.8 \ cm = \frac{8.8 }{10} = 0.88 \ m$

   The velocity at location 1 is  $v_1 = 2.4 \ m /s$

   The diameter at location 2 is  $d_2 = 5.80 \ cm = 0.58 \ m$

Generally the area at location 1 is  

       $A_1 = \pi * \frac{d^2}{ 2}$

=>     $A_1 = \pi * \frac{0.88^2}{ 2}$

=>     $A_1 = 3.142 * \frac{0.88^2}{ 2}$

=>     $A_1 = 1.2166 \ m^2$

Generally the area at location 1 is  

       $A_2 = \pi * \frac{d_1^2}{ 2}$

=>     $A_2 = \pi * \frac{0.58^2}{ 2}$

=>     $A_2 = 0.528 \ m^2$

Generally from continuity equation we have that

     $A_1 * v_1 = A_2 * v_2$

=>   $1.2166 * 2.4 = 0.528 * v_2$

=>   $1.2166 * 2.4 = 0.528 * v_2$

=>    $v_2 = 5.53 \ m /s$